when:
  path: /v1beta/models/gemini-2.5-flash:generateContent
  method: POST
  query_param:
  - name: key
    value: '[REDACTED]'
  header:
  - name: accept
    value: '*/*'
  - name: content-type
    value: application/json
  body: '{"contents":[{"parts":[{"text":"A train leaves Station A at 60 km/h. Another train leaves Station B (300 km away) 30 minutes later at 90 km/h heading toward Station A. At what time do they meet, and how far from Station A? Show your work.","thought":false}],"role":"user"}],"generationConfig":{"thinkingConfig":{"includeThoughts":true,"thinkingBudget":2048}},"safetySettings":null,"systemInstruction":{"parts":[{"text":"You are a helpful math tutor. Be concise.","thought":false}],"role":"model"},"toolConfig":null}'
then:
  status: 200
  header:
  - name: content-type
    value: application/json; charset=UTF-8
  body: '{"candidates":[{"content":{"parts":[{"text":"**Relative Motion Problem: Train Analysis**\n\nOkay, so I''ve got a classic relative motion problem here with these two trains. My immediate goal is to figure out when and where they''ll meet. I need to nail down two key pieces of information: the time they rendezvous and the distance from Station A where that happens.\n\nMy approach is pretty standard. First, I''ll need to define all the variables involved – speeds, distances, times – making sure everything is clearly labeled. Then, I need to account for that crucial 30-minute head start the first train gets. This is key, as it''s not a simultaneous start! Next comes the meat of it: setting up those distance equations based on the relative motion. The sum of the distances traveled should equal the total distance between the stations when they meet. From there, I''ll solve for the meeting time, and finally, calculate that distance from Station A.\n\nLet me lay it out formally. I''m calling the first train''s speed $v_A$, the second train''s speed $v_B$, the total distance between stations $D$. I''ll use $t_A$ and $t_B$ to represent each train''s travel time, and $d_A$ and $d_B$ for their respective distances. $v_A$ is 60 km/h, $v_B$ is 90 km/h, and $D$ is 300 km.\n\nThe first train''s 30-minute lead (0.5 hours) needs careful handling. I''ll define ''t'' as the time elapsed *after* the second train leaves. This means the first train has been running for $t + 0.5$ hours. Now I can build the distance equations:  $d_A = v_A \\times (t + 0.5)$ and $d_B = v_B \\times t$.\n\nWhen they meet, the total distance traveled is D.  Therefore: $d_A + d_B = D$.  Substituting, I get: $v_A \\times (t + 0.5) + v_B \\times t = D$. Plugging in the values I know:  $60 \\times (t + 0.5) + 90 \\times t = 300$. Expanding and simplifying, I get $150t = 270$, which gives me $t = 1.8$ hours. That''s the time elapsed *after* train B leaves.\n\nNow, to get the distances: $d_A = 60 \\times (1.8 + 0.5) = 138$ km and $d_B = 90 \\times 1.8 = 162$ km. Let''s do a quick sanity check: 138 + 162 does indeed equal 300 km. That confirms my calculations.\n\nOkay, so what about that question of *when* they meet? The problem is a little ambiguous here, as I see it. It looks like it wants the clock time. But no initial departure time for the trains is specified. I need to make an assumption. They meet 1.8 hours after B leaves, which is 2.3 hours from the initial departure of A. The question''s wording requires I state *how much time passed from the start* of the first train''s journey. So, if Train A leaves at an arbitrary 0:00, then Train B leaves at 0:30, and the meeting is 2.3 hours after Train A''s departure time, which is 2 hours and 18 minutes.\n\nThey meet 2.3 hours, or 2 hours and 18 minutes after Train A departs, and the meeting point is $d_A = 138$ km from Station A.\n","thought":true},{"text":"Let $t$ be the time in hours that Train B travels until they meet.\nSince Train A leaves 30 minutes (0.5 hours) earlier, Train A travels for $t + 0.5$ hours.\n\nThe total distance between stations A and B is 300 km.\nWhen the trains meet, the sum of the distances they''ve traveled equals the total distance.\n\nDistance traveled by Train A ($d_A$):\n$d_A = \\text{Speed}_A \\times \\text{Time}_A = 60 \\text{ km/h} \\times (t + 0.5) \\text{ h}$\n\nDistance traveled by Train B ($d_B$):\n$d_B = \\text{Speed}_B \\times \\text{Time}_B = 90 \\text{ km/h} \\times t \\text{ h}$\n\nAt the meeting point:\n$d_A + d_B = 300 \\text{ km}$\n$60(t + 0.5) + 90t = 300$\n\nNow, solve for $t$:\n$60t + 30 + 90t = 300$\n$150t + 30 = 300$\n$150t = 300 - 30$\n$150t = 270$\n$t = \\frac{270}{150}$\n$t = \\frac{27}{15}$\n$t = 1.8 \\text{ hours}$\n\nThis is the time Train B travels.\n\n**1. At what time do they meet?**\nThe question asks for \"at what time\", implying a clock time. Since no initial departure time for Train A is given, we will state the duration from when Train A departed.\n\nTrain A travels for $t + 0.5 = 1.8 + 0.5 = 2.3$ hours.\nConvert 2.3 hours to hours and minutes:\n$2.3 \\text{ hours} = 2 \\text{ hours} + (0.3 \\times 60) \\text{ minutes} = 2 \\text{ hours and } 18 \\text{ minutes}$.\n\nThey meet **2 hours and 18 minutes after Train A departed.**\n\n**2. How far from Station A?**\nThis is the distance traveled by Train A:\n$d_A = 60 \\text{ km/h} \\times (2.3 \\text{ h})$\n$d_A = 138 \\text{ km}$\n\nAlternatively, we can check with Train B''s distance:\n$d_B = 90 \\text{ km/h} \\times (1.8 \\text{ h})$\n$d_B = 162 \\text{ km}$\nThe sum is $138 + 162 = 300$ km, which is correct.\nThe meeting point is 138 km from Station A.\n\n**Summary:**\n*   They meet **2 hours and 18 minutes after Train A''s departure**.\n*   They meet **138 km from Station A**."}],"role":"model"},"finishReason":"STOP","index":0}],"modelVersion":"gemini-2.5-flash","responseId":"id_REDACTED_1","usageMetadata":{"candidatesTokenCount":716,"promptTokenCount":72,"promptTokensDetails":[{"modality":"TEXT","tokenCount":72}],"serviceTier":"standard","thoughtsTokenCount":1236,"totalTokenCount":2024}}'
---
when:
  path: /v1beta/models/gemini-2.5-flash:generateContent
  method: POST
  query_param:
  - name: key
    value: '[REDACTED]'
  header:
  - name: accept
    value: '*/*'
  - name: content-type
    value: application/json
  body: '{"contents":[{"parts":[{"text":"A train leaves Station A at 60 km/h. Another train leaves Station B (300 km away) 30 minutes later at 90 km/h heading toward Station A. At what time do they meet, and how far from Station A? Show your work.","thought":false}],"role":"user"},{"parts":[{"text":"**Relative Motion Problem: Train Analysis**\n\nOkay, so I''ve got a classic relative motion problem here with these two trains. My immediate goal is to figure out when and where they''ll meet. I need to nail down two key pieces of information: the time they rendezvous and the distance from Station A where that happens.\n\nMy approach is pretty standard. First, I''ll need to define all the variables involved – speeds, distances, times – making sure everything is clearly labeled. Then, I need to account for that crucial 30-minute head start the first train gets. This is key, as it''s not a simultaneous start! Next comes the meat of it: setting up those distance equations based on the relative motion. The sum of the distances traveled should equal the total distance between the stations when they meet. From there, I''ll solve for the meeting time, and finally, calculate that distance from Station A.\n\nLet me lay it out formally. I''m calling the first train''s speed $v_A$, the second train''s speed $v_B$, the total distance between stations $D$. I''ll use $t_A$ and $t_B$ to represent each train''s travel time, and $d_A$ and $d_B$ for their respective distances. $v_A$ is 60 km/h, $v_B$ is 90 km/h, and $D$ is 300 km.\n\nThe first train''s 30-minute lead (0.5 hours) needs careful handling. I''ll define ''t'' as the time elapsed *after* the second train leaves. This means the first train has been running for $t + 0.5$ hours. Now I can build the distance equations:  $d_A = v_A \\times (t + 0.5)$ and $d_B = v_B \\times t$.\n\nWhen they meet, the total distance traveled is D.  Therefore: $d_A + d_B = D$.  Substituting, I get: $v_A \\times (t + 0.5) + v_B \\times t = D$. Plugging in the values I know:  $60 \\times (t + 0.5) + 90 \\times t = 300$. Expanding and simplifying, I get $150t = 270$, which gives me $t = 1.8$ hours. That''s the time elapsed *after* train B leaves.\n\nNow, to get the distances: $d_A = 60 \\times (1.8 + 0.5) = 138$ km and $d_B = 90 \\times 1.8 = 162$ km. Let''s do a quick sanity check: 138 + 162 does indeed equal 300 km. That confirms my calculations.\n\nOkay, so what about that question of *when* they meet? The problem is a little ambiguous here, as I see it. It looks like it wants the clock time. But no initial departure time for the trains is specified. I need to make an assumption. They meet 1.8 hours after B leaves, which is 2.3 hours from the initial departure of A. The question''s wording requires I state *how much time passed from the start* of the first train''s journey. So, if Train A leaves at an arbitrary 0:00, then Train B leaves at 0:30, and the meeting is 2.3 hours after Train A''s departure time, which is 2 hours and 18 minutes.\n\nThey meet 2.3 hours, or 2 hours and 18 minutes after Train A departs, and the meeting point is $d_A = 138$ km from Station A.\n","thought":true},{"text":"Let $t$ be the time in hours that Train B travels until they meet.\nSince Train A leaves 30 minutes (0.5 hours) earlier, Train A travels for $t + 0.5$ hours.\n\nThe total distance between stations A and B is 300 km.\nWhen the trains meet, the sum of the distances they''ve traveled equals the total distance.\n\nDistance traveled by Train A ($d_A$):\n$d_A = \\text{Speed}_A \\times \\text{Time}_A = 60 \\text{ km/h} \\times (t + 0.5) \\text{ h}$\n\nDistance traveled by Train B ($d_B$):\n$d_B = \\text{Speed}_B \\times \\text{Time}_B = 90 \\text{ km/h} \\times t \\text{ h}$\n\nAt the meeting point:\n$d_A + d_B = 300 \\text{ km}$\n$60(t + 0.5) + 90t = 300$\n\nNow, solve for $t$:\n$60t + 30 + 90t = 300$\n$150t + 30 = 300$\n$150t = 300 - 30$\n$150t = 270$\n$t = \\frac{270}{150}$\n$t = \\frac{27}{15}$\n$t = 1.8 \\text{ hours}$\n\nThis is the time Train B travels.\n\n**1. At what time do they meet?**\nThe question asks for \"at what time\", implying a clock time. Since no initial departure time for Train A is given, we will state the duration from when Train A departed.\n\nTrain A travels for $t + 0.5 = 1.8 + 0.5 = 2.3$ hours.\nConvert 2.3 hours to hours and minutes:\n$2.3 \\text{ hours} = 2 \\text{ hours} + (0.3 \\times 60) \\text{ minutes} = 2 \\text{ hours and } 18 \\text{ minutes}$.\n\nThey meet **2 hours and 18 minutes after Train A departed.**\n\n**2. How far from Station A?**\nThis is the distance traveled by Train A:\n$d_A = 60 \\text{ km/h} \\times (2.3 \\text{ h})$\n$d_A = 138 \\text{ km}$\n\nAlternatively, we can check with Train B''s distance:\n$d_B = 90 \\text{ km/h} \\times (1.8 \\text{ h})$\n$d_B = 162 \\text{ km}$\nThe sum is $138 + 162 = 300$ km, which is correct.\nThe meeting point is 138 km from Station A.\n\n**Summary:**\n*   They meet **2 hours and 18 minutes after Train A''s departure**.\n*   They meet **138 km from Station A**.","thought":false}],"role":"model"},{"parts":[{"text":"Now suppose both trains slow down by 10 km/h after traveling half the original distance. When do they meet now?","thought":false}],"role":"user"}],"generationConfig":{"thinkingConfig":{"includeThoughts":true,"thinkingBudget":2048}},"safetySettings":null,"systemInstruction":{"parts":[{"text":"You are a helpful math tutor. Be concise.","thought":false}],"role":"model"},"toolConfig":null}'
then:
  status: 200
  header:
  - name: content-type
    value: application/json; charset=UTF-8
  body: '{"candidates":[{"content":{"parts":[{"text":"**Analyzing the Modified Train Problem**\n\nOkay, so I''m revisiting the train problem with a new twist. We''ve got the standard setup: two trains, A and B, starting from opposite ends of a 300 km track. Train A starts with a speed of 60 km/h, Train B starts at 90 km/h, and Train B has a 30-minute head start. The crucial change is that *after* each train covers half the original distance (150 km) *individually*, they both reduce their speed by 10 km/h. So, Train A slows to 50 km/h, and Train B slows to 80 km/h after reaching the 150 km mark from their respective starting points.\n\nThis means the solution will involve piecewise functions and different stages of motion, so I should track both time and position for each train. I must find when they meet (time from when train A departs).\n\nThe key is that each train''s speed reduction happens independently, based on *its* own distance traveled. It''s *not* a shared midpoint of the track. This interpretation is critical. I''ve re-read the problem statement to ensure the speeds are changed *after* each train travels half of the original distance.\n\nSo, the meeting point will likely fall into one of a few scenarios. It''s very possible both meet before either of them has reached the 150km mark, as was the case with the previous problem solution. It''s important to clarify the specifics of this new scenario with a new solution. The other possibilities are:\n\n1. Both trains meet before either reaches 150 km. (Unlikely)\n2. Train A hits 150km, slows down, and then they meet. Train B *has not yet* reached 150 km.\n3. Train B hits 150km, slows down, and then they meet. Train A *has not yet* reached 150 km. (Highly unlikely)\n4. Both trains hit 150km, slow down, and *then* they meet.\n\nTo tackle this, let''s calculate when each train reaches the 150 km mark and changes its speed, relative to Train A''s departure.\n\n*   **Train A:** Traveling at 60 km/h, it reaches 150 km after 2.5 hours.\n*   **Train B:** Traveling at 90 km/h, it reaches 150 km after 5/3 hours, or approximately 1.67 hours. However, remember the delay. Train B starts 0.5 hours later. So, relative to Train A''s departure, Train B reaches 150 km after 13/6 hours (approximately 2.17 hours).\n\nSo the speeds are changed at 13/6 hours. At this point, I see that Train A is at 130 km from A, while Train B is 150 km from B (or 150 km from A). The trains are only 20 km apart. Train A''s speed still hasn''t changed, but Train B''s has. I will need to use this information. This setup should give me a good framework to work from in building the model, keeping track of the position and speed changes.\n","thought":true},{"text":"This problem requires tracking the distances and times in stages, as the speeds change.\n\nLet $T$ be the time in hours measured from Train A''s departure.\n\n**1. Determine when each train''s speed changes:**\n\n*   **Train A:** Travels 150 km at 60 km/h.\n    Time taken for Train A to travel 150 km: $t_{A150} = \\frac{150 \\text{ km}}{60 \\text{ km/h}} = 2.5 \\text{ hours}$.\n    So, after $T = 2.5$ hours from Train A''s departure, Train A''s speed changes from 60 km/h to $60 - 10 = 50$ km/h.\n\n*   **Train B:** Travels 150 km at 90 km/h.\n    Time taken for Train B to travel 150 km: $t_{B150} = \\frac{150 \\text{ km}}{90 \\text{ km/h}} = \\frac{5}{3} \\text{ hours} \\approx 1.67 \\text{ hours}$.\n    Train B departs 0.5 hours after Train A. So, relative to Train A''s departure, Train B''s speed changes at $T = 0.5 + \\frac{5}{3} = \\frac{3}{6} + \\frac{10}{6} = \\frac{13}{6} \\text{ hours}$.\n    $\\frac{13}{6} \\text{ hours} \\approx 2.17 \\text{ hours}$.\n    At this time, Train B''s speed changes from 90 km/h to $90 - 10 = 80$ km/h.\n\n**2. Analyze the meeting scenario:**\n\nNotice that Train B changes speed at $T = \\frac{13}{6}$ hours, and Train A changes speed at $T = 2.5 = \\frac{15}{6}$ hours. Since $\\frac{13}{6} < \\frac{15}{6}$, Train B will change its speed *before* Train A changes its speed.\n\n**Let''s analyze the movement up to $T = \\frac{13}{6}$ hours (when Train B changes speed):**\n\n*   **Train A''s distance:** $d_A = 60 \\text{ km/h} \\times \\frac{13}{6} \\text{ h} = 130 \\text{ km}$ from Station A.\n*   **Train B''s distance:** Train B has traveled 150 km from Station B, so it is $300 - 150 = 150 \\text{ km}$ from Station A.\n*   **Distance between them:** At $T = \\frac{13}{6}$ hours, the distance separating them is $150 \\text{ km} - 130 \\text{ km} = 20 \\text{ km}$.\n*   **Speeds at this point:** Train A is still at 60 km/h. Train B has just slowed down to 80 km/h.\n\n**3. Calculate the time to meet for the remaining distance:**\n\nNow we have a remaining distance of 20 km between them. Train A is at 130 km from A, moving towards B at 60 km/h. Train B is at 150 km from A, moving towards A at 80 km/h.\nBoth trains are moving towards each other.\nTheir relative speed is $60 \\text{ km/h} + 80 \\text{ km/h} = 140 \\text{ km/h}$.\n\nTime to meet from this point ($t_{meet\\_rem}$):\n$t_{meet\\_rem} = \\frac{20 \\text{ km}}{140 \\text{ km/h}} = \\frac{1}{7} \\text{ hours}$.\n\n**4. Total time until they meet:**\n\nThe meeting occurs $\\frac{1}{7}$ hours after $T = \\frac{13}{6}$ hours.\nTotal time from Train A''s departure ($T_{total}$):\n$T_{total} = \\frac{13}{6} \\text{ h} + \\frac{1}{7} \\text{ h}$\n$T_{total} = \\frac{13 \\times 7}{42} + \\frac{1 \\times 6}{42} = \\frac{91 + 6}{42} = \\frac{97}{42} \\text{ hours}$.\n\nConvert to hours and minutes:\n$\\frac{97}{42} \\text{ hours} \\approx 2.3095 \\text{ hours}$.\n$2 \\text{ hours } + (0.3095 \\times 60) \\text{ minutes } \\approx 2 \\text{ hours and } 18.57 \\text{ minutes}$.\nLet''s use fractions for precision.\n$T_{total} = 2 \\text{ hours } \\frac{13}{42} \\text{ hours}$.\n$\\frac{13}{42} \\times 60 \\text{ minutes} = \\frac{13 \\times 10}{7} = \\frac{130}{7} \\text{ minutes} \\approx 18.57 \\text{ minutes}$.\n\nThey meet **approximately 2 hours and 18.6 minutes after Train A departed.**\n\n**5. How far from Station A?**\n\nThis is the distance Train A travels in $T_{total}$.\nFor the first $\\frac{13}{6}$ hours, Train A travels at 60 km/h.\nDistance A in stage 1 ($d_{A1}$) = $60 \\text{ km/h} \\times \\frac{13}{6} \\text{ h} = 130 \\text{ km}$.\n\nFor the remaining $\\frac{1}{7}$ hours, Train A continues at 60 km/h because it has not reached its 150 km mark yet. (This is important! Our analysis confirmed Train A does not slow down before meeting).\nDistance A in stage 2 ($d_{A2}$) = $60 \\text{ km/h} \\times \\frac{1}{7} \\text{ h} = \\frac{60}{7} \\text{ km} \\approx 8.57 \\text{ km}$.\n\nTotal distance from Station A ($D_A$):\n$D_A = 130 \\text{ km} + \\frac{60}{7} \\text{ km} = \\frac{910}{7} + \\frac{60}{7} = \\frac{970}{7} \\text{ km}$.\n$\\frac{970}{7} \\text{ km} \\approx 138.57 \\text{ km}$.\n\n**Check using Train B''s distance:**\nTrain B travels for $\\frac{1}{7}$ hours (at 80 km/h) *after* traveling 150 km.\nDistance B in stage 1 ($d_{B1}$) = 150 km.\nDistance B in stage 2 ($d_{B2}$) = $80 \\text{ km/h} \\times \\frac{1}{7} \\text{ h} = \\frac{80}{7} \\text{ km}$.\nTotal distance from Station B ($D_B$) = $150 + \\frac{80}{7} = \\frac{1050 + 80}{7} = \\frac{1130}{7} \\text{ km}$.\nDistance from Station A = $300 - D_B = 300 - \\frac{1130}{7} = \\frac{2100 - 1130}{7} = \\frac{970}{7} \\text{ km}$. This matches $D_A$.\n\n**Summary:**\n*   They meet **$\\frac{97}{42}$ hours (or approximately 2 hours and 18.6 minutes) after Train A''s departure.**\n*   They meet **$\\frac{970}{7}$ km (or approximately 138.57 km) from Station A.**"}],"role":"model"},"finishReason":"STOP","index":0}],"modelVersion":"gemini-2.5-flash","responseId":"id_REDACTED_2","usageMetadata":{"candidatesTokenCount":1874,"promptTokenCount":1654,"promptTokensDetails":[{"modality":"TEXT","tokenCount":1654}],"serviceTier":"standard","thoughtsTokenCount":1453,"totalTokenCount":4981}}'
